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Bowl and Rod (Posted on 2004-12-31) Difficulty: 3 of 5
Given a solid bowl (an idealized hemisphere, the opening facing straight up), let a slender rod rest in the bowl supported at two points: a point on the bowl's edge, and some point on the bowl's interior surface.

The bowl's edge exerts a force on the rod which is perpendicular to the rod's length.

The bowl's interior exerts, on the rod's end, a force which is perpendicular to the bowl's surface (at the point where they meet).

If the rod's length is three times the radius of the bowl, what is the angle between the rod, and the plane of the bowl's edge?

See The Solution Submitted by Bractals    
Rating: 2.0000 (2 votes)

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Solution Solution | Comment 7 of 18 |

Finally got it, I think.  See my first post for definitions.

Final answer, with the help of Excel "goal seek":
Angle t is: .255673 radians, or 14.649 degrees.

Eqn 2: mg sin(t) = FI cos(t)
Eqn 3: mg (L/3r) cos(t) (L/2) + FI sin(t) (3r-L) = mg ((3r-L)/3r) cos(t) (3r-L)/2
Eqn 4: 7r^2 - 6rL + L^2 = 2r^2 cos(2t)

Solve Eqn 2 for FI, then plug in to Eqn 3, divide out mg
   FI = mg tan(t)
Eqn 3':  (L/3r) cos(t) (L/2) + tan(t) sin(t) (3r-L) =  ((3r-L)/3r) cos(t) (3r-L)/2 which can be solved for L
    (fortunately, some L^2 cos(t) terms cancel out)
    { note that cos(2x)= cos^2(x) - sin^2(x) }

Eqn 3'':   L = (3r/2)* {1 - [sin(t)/cos(2t)]}
Now plug this value for L into Eqn 4
     { both Eqn's 3 and 4 are completely in terms of r, L, and t  but r isn't a variable, it's a parameter so we have 2 equations and 2 unknowns }

let f(t)=sin(t)/cos(2t) for ease of writing
after lots of messy math, I simplified Eqn 4 to:

8 cos(2t) - 18 f(t) -9 [f(t)]^2 =1
Which I plugged into Excel's goal seek solver and obtained the answer above.
I may very well have made a math error.   Hope someone can check me on this.  Seems like there should be an easier way.


  Posted by Larry on 2005-01-01 14:03:19
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